Two parallel plate air capacitors,each of capacitance X F are in series with a battery of 12-V.If a dielectric with $\kappa$ = 3 is inserted between the plates of one of the capacitors,then calculate the change in electrical charge (in Coulombs)that occurs on one of its plates.
A)9X
B)3X/4
C)3X
D)16X
E)none of the above

Question

Quizplus · Accepted Answer

When a dielectric is inserted between the plates of a capacitor, the capacitance increases by a factor of the dielectric constant. In this case, the capacitance of one of the capacitors increases to F1F1F1S13X= 3X.

Since the capacitors are in series, the charge on both capacitors is the same. Let Q be the charge on each capacitor. The equivalent capacitance of the series combination is given by:

Ceq = (C1*C2)/(C1+C2)

where C1 = C2 = X and C2 has been replaced with 3X due to the insertion of the dielectric.

Plugging in the values, we get:

Ceq = (X*3X)/(X+3X) = 3X/4

The potential difference across the capacitors is 12V. Therefore, the charge on each capacitor is given by:

Q = Ceq*V

Q = (3X/4)*12 = 9X

Hence, the change in charge on one of the plates is 9X - X = 8X. Therefore, option C is the correct choice.

Two Parallel Plate Air Capacitors,each of Capacitance X F Are κ\kappaκ

Two Parallel Plate Air Capacitors,each of Capacitance X F Are $\kappa$